A long solenoid has `1000` turns. When a current of `4A` flows through it, the magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb`. The self-inductance of the solenoid is

Options

A. 3 H
B. 2 H
C. 1 H
D. 4 H

Answer

C

Explanation

Given, number of turns of solenoid, N = 1000
Current, I = 4A
Magnetic flux, `phi_(B)=4xx10^(-3)Wb`
`because` Self-inductance of solenoid is given by
`L=(phi_(B)N)/(I)`
Substituting the given values in Eq. (i), we get
`L=(4xx10^(-3)xx1000)/(4)=1H`